Mrs. Tandon has two sons, one being exactly one year older than the other. At present, her age is equal to the sum of the squares of the ages of her sons. If 4 years hence her age becomes five times the age of the elder son then find the present ages of her sons.
Ans.=4 years, 5years as given in textbook.
Solve it as a quadratic equation.
Answers
Then age of elder son=x+1
Given that present age of Mrs. Tandon is equal to sum of the squares of two sons.
Then,
Present age of Mrs.Tandon=x²+(x+1)²
Present age of Mrs.Tandon=2x²+2x+1
Age of Mrs.Tandon after 4 years=2x²+2x+1+4=2x²+2x+5
Age of elder son after 4 years=x+1+4=x+5
As per given condition we have,
2x²+2x+5=5(x+5)
2x²+2x+5=5x+25
2x²-3x-20=0
2x²-8x+5x-20=0
2x(x-4)+5(x-4)=0
(2x+5)(x-4)=0
x=-5/2 (or) x=4
Here x belongs to Natural numbers.
Therefore x=4
Hence,
present age of younger son=x=4years
present age of elder son=x+1=5years.
Answer:
Let , the age of elder son be "x".
Then the age younger son = x -1.
given that,
Tendon 's age is equal to the sum of
squares of her two son's.
So, tendon 's age = (x)² +(x - 1)²
= x² + x² + 1² - 2x
Tendon 's age = (2x² - 2x + 1 .) years.-- eq-(1)
And also given that, in 4 years her age will be 5 times the age of the eldest son.
After 4 years , Elder son's age = x + 4.
Tendon's age = 5(x + 4)
Tendon's age = (5x + 20) years.-- eq-(2)
From eq -(1) ,
Tendon's age after 4 years = (2x² - 2x +1 ) + 4
Tendon's age = 2x²- 2x +5 -- eq- (3)
Equate eq (2) & (3).
5x + 20 = 2x² - 2x + 5
2x² - 2x +5 - 5x - 20 = 0
2x² - 7x - 15 = 0
2x² - 10x + 3x - 15 = 0
2x ( x - 5) + 3(x - 5) = 0
(2x + 3)(x - 5) = 0
2x + 3 = 0
2x = -3
x = -3/2.
x - 5 = 0
x = 5.
Age of a person cannot be negative so Positive value is taken.
Hence, x = 5.
Therefore, Age of elder son = x = 5 years and age of younger son = (x - 1) = (5-1) = 4 years.
Step-by-step explanation: